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3 years ago

The table below represents an exponential function. what is the missing value?

Mathematics

1 answer:

Blababa [14]3 years ago

4 0

Answer:

The missing value is 128

Step-by-step explanation:

we know that

The equation of the exponential function is of the form

$f(x)=a(b^{x})$

where

a is the initial value (value of the function when the value of x is equal to zero)

b is the base

r is the rate

b=(1+r)

In this problem , looking at the table, we have that the number of bacteria is doubled every second

$r=100\%=100/100=1\\b=1+r=1+1=2\\a=8$

substitute in the formula

$f(x)=8(2^{x})$

where

f(x) is the number of bacteria

x is the time in seconds

For x=4 sec

substitute

$f(4)=8(2^{4})$

$f(4)=128\ bacteria$

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Gekata [30.6K]

Answer:

negative

Step-by-step explanation:

okay so look at the rise and the run of this line (that is given). slope is defined as rise divided by run. the "rise" is the change in y values, while the "run" is the change in x values. the rise in this problem is -4, as shown on the graph. the run is 3, as also shown there. therefore, the slope is -4/3. the type of slope this is is negative slope because the slope is clearly negative.

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4 years ago

Read 2 more answers

Four glasses of milk and 3 snack bars have a total of 82 carbohydrates (carbs), and 2 glasses of milk and 4 snack bars have a to

vredina [299]

Answer:

milk: 10 carbs
snack bar: 14 carbs

Step-by-step explanation:

Let m and s represent the carbs in the milk and snack bar, respectively. The given relations are ...

4m +3s = 82

2m +4s = 76

Subtracting the first equation from twice the second gives ...

2(2m +4s) -(4m +3s) = 2(76) -(82)

5s = 70 . . . . . . . simplify

s = 14 . . . . . . . divide by 5

We substitute this into the second equation to find m.

2m +4(14) = 76

m +28 = 38 . . . . divide by 2

m = 10 . . . . . . subtract 28

There are 10 carbs in a glass of milk; 14 carbs in a snack bar.

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2 years ago

The probability that a lab specimen contains high levels of contamination is 0.10. Five samples are checked, and the samples are

raketka [301]

Answer:

(a) 0.59049 (b) 0.32805 (c) 0.40951

Step-by-step explanation:

Let's define

$A_{i}$ : the lab specimen number i contains high levels of contamination for i = 1, 2, 3, 4, 5, so,

$P(A_{i})=0.1$ for i = 1, 2, 3, 4, 5

The complement for $A_{i}$ is given by

$A_{i}^{$c$}$ : the lab specimen number i does not contains high levels of contamination for i = 1, 2, 3, 4, 5, so

P( $A_{i}^{$c$}$ )=0.9 for i = 1, 2, 3, 4, 5

(a) The probability that none contain high levels of contamination is given by

P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )= $(0.9)^{5}$ =0.59049 because we have independent events.

(b) The probability that exactly one contains high levels of contamination is given by

P( $A_{1}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}$ )=5×(0.1)× $(0.9)^{4}$ =0.32805

because we have independent events.

P( $A_{1}$ ∪ $A_{2}$ ∪ $A_{3}$ ∪ $A_{4}$ ∪ $A_{5}$ )=1-P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )=1-0.59049=0.40951

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